∫ x^2/(1+x^3)^2dx=?以及∫ x+1/(x^2+2x-3)^2dx=?我看懂了,你是把括号里的东西用y来代替,不过你第二题的答案和我书本后面的答案不一样,我书本后面的答案是 -1/2y +C
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![∫ x^2/(1+x^3)^2dx=?以及∫ x+1/(x^2+2x-3)^2dx=?我看懂了,你是把括号里的东西用y来代替,不过你第二题的答案和我书本后面的答案不一样,我书本后面的答案是 -1/2y +C](/uploads/image/z/14867486-62-6.jpg?t=%E2%88%AB+x%5E2%2F%281%2Bx%5E3%29%5E2dx%3D%3F%E4%BB%A5%E5%8F%8A%E2%88%AB+x%2B1%2F%28x%5E2%2B2x-3%29%5E2dx%3D%3F%E6%88%91%E7%9C%8B%E6%87%82%E4%BA%86%EF%BC%8C%E4%BD%A0%E6%98%AF%E6%8A%8A%E6%8B%AC%E5%8F%B7%E9%87%8C%E7%9A%84%E4%B8%9C%E8%A5%BF%E7%94%A8y%E6%9D%A5%E4%BB%A3%E6%9B%BF%EF%BC%8C%E4%B8%8D%E8%BF%87%E4%BD%A0%E7%AC%AC%E4%BA%8C%E9%A2%98%E7%9A%84%E7%AD%94%E6%A1%88%E5%92%8C%E6%88%91%E4%B9%A6%E6%9C%AC%E5%90%8E%E9%9D%A2%E7%9A%84%E7%AD%94%E6%A1%88%E4%B8%8D%E4%B8%80%E6%A0%B7%EF%BC%8C%E6%88%91%E4%B9%A6%E6%9C%AC%E5%90%8E%E9%9D%A2%E7%9A%84%E7%AD%94%E6%A1%88%E6%98%AF+-1%2F2y+%2BC)
∫ x^2/(1+x^3)^2dx=?以及∫ x+1/(x^2+2x-3)^2dx=?我看懂了,你是把括号里的东西用y来代替,不过你第二题的答案和我书本后面的答案不一样,我书本后面的答案是 -1/2y +C
∫ x^2/(1+x^3)^2dx=?以及∫ x+1/(x^2+2x-3)^2dx=?
我看懂了,你是把括号里的东西用y来代替,不过你第二题的答案和我书本后面的答案不一样,我书本后面的答案是 -1/2y +C
∫ x^2/(1+x^3)^2dx=?以及∫ x+1/(x^2+2x-3)^2dx=?我看懂了,你是把括号里的东西用y来代替,不过你第二题的答案和我书本后面的答案不一样,我书本后面的答案是 -1/2y +C
对,用基础方法解可以了,不用太复杂,2条的解法类似
ln|tan(t/2)|+c
1、∫ x^2/(1+x^3)^2dx
采用凑微分法,注意到(x^3)’=3x^2,所以
∫ x^2/(1+x^3)^2dx = (1/3)∫ 1/(1+x^3)^2d(x^3) = (1/3)∫ 1/(1+x^3)^2d(x^3 + 1)
= (1/3)∫ 1/y^2 dy = -1/(3y) + C
2、∫ x+1/(x^2+2x-3)^2dx
∫ x...
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1、∫ x^2/(1+x^3)^2dx
采用凑微分法,注意到(x^3)’=3x^2,所以
∫ x^2/(1+x^3)^2dx = (1/3)∫ 1/(1+x^3)^2d(x^3) = (1/3)∫ 1/(1+x^3)^2d(x^3 + 1)
= (1/3)∫ 1/y^2 dy = -1/(3y) + C
2、∫ x+1/(x^2+2x-3)^2dx
∫ x+1/(x^2+2x-3)^2dx = ∫ xdx + ∫1/(x^2+2x-3)^2dx
=(x^2)/2 + ∫1/(x-1)^2(x+3)^2dx :把(1/(x-1)(x+3))^2拆成 (1/4)*(1/(x-1) - 1/(x+3))^2
=(x^2)/2 + (1/4)∫1/(x-1)^2dx + (2/4)∫1/(x-1)(x+3)dx + (1/4)∫1/(x+3)^2dx
=(x^2)/2 + (1/4)∫1/(x-1)^2d(x-1) + (2/4)(1/4){∫1/(x-1)dx - ∫1/(x+3)dx}+ (1/4)∫1/(x+3)^2d(x+3)
到这一步已经拆成基本函数了,后面记不清原函数是否能直接合并,我是按照合并处理的
=(x^2)/2 -1/(4y) + (1/8)lnx - (1/8)lnx - 1/(4y) + C
= (x^2)/2 -1/(2y) + C
这个是最基础的方法,第一类换元法,我可以保证不会有更简单的方法了,你是一点儿也看不懂?
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